A rod of a certain metal is 1.0 , m long and | vedclass

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(B) The resistivity $ho$ of the rod material is given by $ho = \frac{RA}{l}$.Given: $R = 3.0 	imes 10^{-3} \, \Omega$,$l = 1.0 \, m$,and diameter

Vedclass · Accepted Answer

$2.70 	imes 10^{-7} \, \Omega$

Vedclass · Answer

(B) The resistivity $ho$ of the rod material is given by $ho = \frac{RA}{l}$.Given: $R = 3.0 	imes 10^{-3} \, \Omega$,$l = 1.0 \, m$,and diameter $d_1 = 0.6 \, cm = 0.6 	imes 10^{-2} \, m$. The radius $r_1 = 0.3 	imes 10^{-2} \, m$.$ho = \frac{3.0 	imes 10^{-3} 	imes \pi 	imes (0.3 	imes 10^{-2})^2}{1.0} = 3.0 	imes 10^{-3} 	imes \pi 	imes 0.09 	imes 10^{-4} = 27 	imes 10^{-9} \pi \, \Omega \cdot m$.For the disc,thickness $t = 1.0 \, mm = 1.0 	imes 10^{-3} \, m$ and diameter $d_2 = 2.0 \, cm = 2.0 	imes 10^{-2} \, m$. The radius $r_2 = 1.0 	imes 10^{-2} \, m$.The resistance $R'$ between the round faces is $R' = ho \frac{t}{A_2} = ho \frac{t}{\pi r_2^2}$.$R' = (27 	imes 10^{-9} \pi) 	imes \frac{1.0 	imes 10^{-3}}{\pi 	imes (1.0 	imes 10^{-2})^2} = \frac{27 	imes 10^{-12}}{10^{-4}} = 27 	imes 10^{-8} = 2.70 	imes 10^{-7} \, \Omega$.

$A$ rod of a certain metal is $1.0 \, m$ long and $0.6 \, cm$ in diameter. Its resistance is $3.0 \times 10^{-3} \, \Omega$. Another disc made of the same metal is $2.0 \, cm$ in diameter and $1.0 \, mm$ thick. What is the resistance between the round faces of the disc?

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